What is the solution for (3y^{''}+2y^'+y)/4 =0 ?

The solution for (3y^{''}+2y^'+y)/4 =0 is y=e^{-t/3}(c_{1}cos((sqrt(2)t)/3)+c_{2}sin((sqrt(2)t)/3))

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(3y^{''}+2y^'+y)/4 =0

\frac{3 y ^{^{''}} + 2 y ^{^{'}} + y}{4} = 0

y = e^{- \frac{t}{3}} (c_{1} cos (\frac{2 t}{3}) + c_{2} sin (\frac{2 t}{3}))

Solution steps

\frac{3 y ^{''} ( t ) + 2 y ^{'} ( t ) + y}{4} = 0

Solve linear ODE:

y = e^{- \frac{t}{3}} (c_{1} cos (\frac{2 t}{3}) + c_{2} sin (\frac{2 t}{3}))

y = e^{- \frac{t}{3}} (c_{1} cos (\frac{2 t}{3}) + c_{2} sin (\frac{2 t}{3}))

16 y^{^{''}} + y = 0

4 y^{^{''}} - 4 y^{^{'}} + 37 y = 0, y (0) = 3, y^{^{'}} (0) = 1.5

y^{^{''}} + 6 y^{^{'}} + 5 y = 0, y (0) = 3, y^{^{'}} (0) = - 1

y^{^{''}} + 3 y^{^{'}} + 8 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 14 y^{^{'}} + 49 y = 0

What is the solution for (3y^{''}+2y^'+y)/4 =0 ?
The solution for (3y^{''}+2y^'+y)/4 =0 is y=e^{-t/3}(c_{1}cos((sqrt(2)t)/3)+c_{2}sin((sqrt(2)t)/3))